What Is the Initial Speed and Maximum Height of a Rock Thrown Vertically?

AI Thread Summary
The discussion focuses on calculating the initial speed and maximum height of a rock thrown vertically using the work-energy theorem and kinematic equations. The rock, weighing 20.4 N, is observed at a height of 14.9 m with a speed of 26.0 m/s. The initial speed is calculated to be approximately 31.11 m/s using the kinematic equation. The mass of the rock is determined to be 2.08 kg, and there is confusion regarding the application of work-energy principles to find the maximum height. Ultimately, the work done by gravity is calculated, which is essential for determining the total energy changes involved in the rock's motion.
Heat
Messages
272
Reaction score
0

Homework Statement



You throw a rock of weight 20.4 N vertically into the air from ground level. You observe that when it is a height 14.9 m above the ground, it is traveling at a speed of 26.0 m/s upward.

Use the work-energy theorem to find its speed just as it left the ground;
Use the work-energy theorem to find its maximum height.
Take the free fall acceleration to be g = 9.80 m/s^2.


Homework Equations


Wtotal = delta K

K = mv^2

v^2 = vi^2 + 2a(x-xi)

The Attempt at a Solution




I believe this can also be done using kinematic equation, since we are given vf,x,a, and we want vi.

So if this is true, I will attempt both ways :D


26^2 = vi^2 + 2(-9.8)(14.9-0)
vi = 31.11 <--------> which is right o:)

now, for using the work energy theorem.

we first need the mass.

using
w=mg

20.4 = m ( 9.8)

m = 2.08

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2

K2 = 843.65
K1 = 0

...how do I proceed?
 
Physics news on Phys.org
since there is no external force (such as friction or air resistance) there is no external work done on theb all except for gravity. So the total initial energy of the ball is equal to the total final energy of the ball

what is the initial energy of the ball and the final energy of the ball?
 
ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?
 
Heat said:
ummm...

so Winitial = Wfinal.

W=Fs

W = (F)(14.9)...?

No, it's not "work initial" = "work final", which in itself would not make sense.

you may either use

change of kinetic energy = total work done by gravity

OR

total final energy = total initial energy

where total energy = kinetic energy plus gravitational potential energy
 
delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2
 
Heat said:
delta k = wg ?

if so, how do I apply this to my problem, as I have this:

K1 = 1.2(2.08)(v)^2
K2 = 1.2(2.08)(26)^2
ok so u have the final and initial kinetic energies
now how do you calculate the work done by gravity??
 
w done by gravity is:

w = mgsin(theta)

w = 2.08 (9.8)(sin 90)
w = 20.38
 
Back
Top